Question

In the figure shown, there is a large sheet of charge of uniform surface charge density $\sigma$. A charge particle of charge $-q$ and mass $m$ is projected from a point $A$ on the sheet with a speed $u$ with angle of projection such that it lands at maximum distance from $A$ on the sheet. Neglecting gravity, find the time of flight.

• A. $\frac{2 \sqrt{2} m v \varepsilon_{0}}{\sigma q}$
• B. $\frac{\sqrt{2} m v \varepsilon_{0}}{5 \sigma q}$
• C. $\frac{m v \varepsilon_{0}}{\sqrt{2} \sigma q}$
• D. $\frac{m \nu \varepsilon_{0}}{2 \sqrt{2} \sigma q}$

Answer 1

Answer: (A)

For maximum distance, angle of projection should be $45^{\circ}$

For vertical displacement: $\vec{S}=\vec{u} t+\frac{1}{2} \vec{a} t^{2}$
$0=u \sin 45^{\circ} t+\frac{1}{2} \cdot \frac{q \cdot \frac{\sigma}{2 \varepsilon_{0}}}{m} t^{2}$
$\Rightarrow t=0$ or $\frac{2 \sqrt{2} m v \varepsilon_{0}}{\sigma q} t=\frac{2 \sqrt{2} m v \varepsilon_{0}}{\sigma q}$ so
$\Rightarrow $ time of flight $=\frac{2 \sqrt{2} m v \varepsilon_{0}}{\sigma q}$